Normality = equivalents/liter of solution.
I think 1 L has a mass of 1.190 kg (or 1190 grams), not 1190 kg.
Of that 1190, how much is HCl.
1190 g x 0.37 = y grams HCl
How many equivalents is that?
y/equivalent mass (and in this case, equivalent mass and molar mass are the same) so y/approx 36.5 (but you do it exactly) = z so the normality is z/1 L = same as z normal which should be about 12 N.
Now just use the dilution formula.
12N*volume = 0.1N*volume you want to prepare.